Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=t^2+t$ and $y=3t^4-t^3$. What is the magnitude of the particle's acceleration vector at $t=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $2$ (Choice C) C $3$ (Choice D) D $4\sqrt{3}$
Answer: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=t^2+t$ and $y=3t^4-t^3$, which means its position vector is $(t^2+t, 3t^4-t^3)$. We are asked to find the magnitude of the particle's acceleration vector at $t=0$. In other words, we need to find $||\vec{a}(0)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^2+t),\dfrac{d}{dt}(3t^4-t^3)\right) \\\\ &=(2t+1,12t^3-3t^2) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(2t+1),\dfrac{d}{dt}(12t^3-3t^2)\right) \\\\ &=(2,36t^2-6t) \end{aligned}$ Finding $\vec{a}(0)$ $\begin{aligned} \vec{a}({0})&=(2,36({0})^2-6({0})) \\\\ &=(2,0-0) \\\\ &=(2,0) \end{aligned}$ Finding $||\vec{a}(0)||$ $\begin{aligned} ||\vec{a}(1)||&=||(C{2},{0})|| \\\\ &=\sqrt{(C{2})^2+({0})^2} \\\\ &=\sqrt{4} \\\\ &=2 \end{aligned}$ In conclusion, the magnitude of the particle's acceleration vector at $t=0$ is $2$.